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Triangles
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
As PR > PQ,
∴ ∠PQR > ∠PRQ (Angle opposite to larger side is larger) … (1)
PS is the bisector of ∠QPR.
∴∠QPS = ∠RPS … (2)
∠PSR is the exterior angle of ΔPQS.
∴ ∠PSR = ∠PQR + ∠QPS … (3)
∠PSQ is the exterior angle of ΔPRS.
∴ ∠PSQ = ∠PRQ + ∠RPS … (4)
Adding equations (1) and (2), we obtain
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [Using the values of equations (3) and (4)]
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In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ΔABC ≅ ΔCDA.
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE
In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
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